83(2009) 313-335
A PROOF OF THE FABER INTERSECTION NUMBER CONJECTURE
Kefeng Liu & Hao Xu
Abstract
We prove the famous Faber intersection number conjecture and other more general results by using a recursion formula ofn-point functions for intersection numbers on moduli spaces of curves. We also present some vanishing properties of Gromov-Witten invari- ants.
1. Introduction
Starting from the work of Mumford, one fundamental problem in algebraic geometry is the study of intersection theory on moduli spaces of stable curves. Through the work of Witten and Kontsevich we learned that the intersection theory of moduli spaces also has striking connection to string theory and two dimensional gravity. Denote by Mg,n the moduli space of stable n-pointed genusg complex algebraic curves. We have the morphism that forgets the last marked point
π:Mg,n+1 −→ Mg,n.
Denote by σ1, . . . , σn the canonical sections of π, and by D1, . . . , Dn the corresponding divisors in Mg,n+1. Let ωπ be the relative dualizing sheaf, we have the following tautological classes on moduli spaces of curves.
ψi =c1(σ∗i(ωπ)) κi =π∗
c1
ωπX
Dii+1 λk=ck(E), 1≤k≤g,
where E=π∗(ωπ) is the Hodge bundle.
Intuitively,ψiis the first Chern class of the line bundle corresponding to the cotangent space of the universal curve at the i-th marked point and the fiber ofEis the space of holomorphic one forms on the algebraic curve.
The classes κi were first introduced by Mumford [22] on Mg, their generalization to Mg,n here is due to Arbarello-Cornalba [1].
Received 10/29/2008.
313
We use Witten’s notation
hτd1· · ·τdnκa1· · ·κam |λk11· · ·λkggi ,
Z
Mg,n
ψ1d1· · ·ψdnnκa1· · ·κamλk11· · ·λkgg.
These intersection numbers are called the Hodge integrals. They are rational numbers because the moduli space of curves are orbifolds (with quotient singularities) except in genus zero. Their degrees should add up to dimMg,n= 3g−3 +n.
Intersection numbers of pureψclasseshτd1· · ·τdniare often called in- tersection indices or descendant integrals. Faber’s algorithm [3] reduces the calculation of general Hodge integrals to intersection indices, based on Mumford’s Chern character formula [22]
ch2g−1(E) = B2g (2g)!
"
κ2g−1−
n
X
i=1
ψi2g−1
+1 2
X
ξ∈∆
lξ∗
2g−2
X
i=0
ψin+1(−ψn+2)2g−2−i
! # , where ∆ enumerates all boundary divisors and lξ∗ is the push-forward map under the natural inclusion.
The celebrated Witten-Kontsevich theorem [13, 25] asserts that the generating function of intersection indices
F(t0, t1, . . .) =X
g
X
n
h
∞
Y
i=0
τiniig
∞
Y
i=0
tnii ni!
is governed by the KdV hierarchy, which provides a recursive way to compute all these intersection numbers.
The tautological ringR∗(Mg) is defined to be the smallestQ-subalge- bra of the Chow ring A∗(Mg) generated by the tautological classes κi
andλi. Mumford [22] proved that the ringR∗(Mg) is in fact generated by the g−2 classesκ1, . . . , κg−2.
It is a theorem of Looijenga [19] that dimRk(Mg) = 0, k > g−2 and dimRg−2(Mg)≤1. Later Faber proved that actually dimRg−2(Mg) = 1.
Faber’s conjecture.Around 1993, Faber [2] proposed three remark- able conjectures about the structure of the tautological ring R∗(Mg) which we briefly state as follows:
i) For 0≤k≤g−2, the natural product
Rk(Mg)×Rg−2−k(Mg)→Rg−2(Mg)∼=Q is a perfect pairing.
ii) The [g/3] classes κ1, . . . , κ[g/3] generate the ringR∗(Mg), with no relations in degrees ≤[g/3].
iii) LetPn
j=1dj =g−2 and dj ≥0. Then (1) π∗(ψ1d1+1. . . ψndn+1) = X
σ∈Sn
κσ = (2g−3 +n)!
(2g−2)!!Qn
j=1(2dj + 1)!!κg−2, whereκσ is defined as follows: write the permutationσ as a prod- uct of ν(σ) disjoint cycles σ =β1· · ·βν(σ), where we think of the symmetric group Sn as acting on the n-tuple (d1, . . . , dn). De- note by |β| the sum of the elements of a cycle β. Then κσ = κ|β1|κ|β2|. . . κ|βν(σ)|.
Part (i) is called Faber’s perfect pairing conjecture, which is still open.
Faber has verified it forg≤23.
Part (ii) has been proved independently by Morita [21] and Ionel [12] with very different methods. As pointed out by Faber [2], Harer’s stability result implies that there is no relation in degrees ≤[g/3].
Part (iii) of Faber’s conjectures is the intersection number conjecture, whose importance lies in that it computes all top intersections in the tautological ringR∗(Mg) and determines its ring structure if we assume Faber’s perfect pairing conjecture. Theoretically it gives the dimension of tautological rings by computing the rank of intersection matrices which we will discuss in a subsequent work.
Faber’s conjecture is a fundamental question mentioned in mono- graphs such as [6,11] that many algebraic geometers have worked on.
In this paper, we prove the Faber intersection number conjecture com- pletely. First we recall two equivalent formulations.
The Faber intersection number conjecture is equivalent to (2)
Z
Mg,n
ψ1d1. . . ψdnnλgλg−1= (2g−3 +n)!|B2g| 22g−1(2g)!Qn
j=1(2dj −1)!!,
where B2g denotes the 2g-th Bernoulli number. By Mumford’s for- mula for the Chern character of the Hodge bundle, the above identity is equivalent to
(3) (2g−3 +n)!
22g−1(2g−1)!Qn
j=1(2dj−1)!! = hτ2g
n
Y
j=1
τdjig−
n
X
j=1
hτdj+2g−1
Y
i6=j
τdiig+1 2
2g−2
X
j=0
(−1)jhτ2g−2−jτj
n
Y
i=1
τdiig−1
+ 1 2
X
n=I‘ J
2g−2
X
j=0
(−1)jhτj
Y
i∈I
τdiig′hτ2g−2−j
Y
i∈J
τdiig−g′,
where dj ≥1,Pn
j=1dj =g+n−2. We refer to [2, 17] for discussions of the above equivalences.
The following interesting relation is observed by Faber and proved by Zagier using the Faber intersection number conjecture (see [2])
κg−21 = 1
g−122g−5((g−2)!)2κg−2.
In fact, from (1), the above relation is equivalent to a combinatorial identity
g
X
k=1
(−1)k
k! (2g+ 1 +k) X
g=m1+···+mk mi>0
2g+k
2m1+ 1, . . . ,2mk+ 1
= (−1)g22g(g!)2. We learned of an elegant proof from Jian Zhou using the residue theo- rem.
Faber [2] proved identity (3) whenn= 1 using explicit formulae of up to three-point functions. The identity (2) was shown to follow from the degree 0 Virasoro conjecture for P2 by Getzler and Pandharipande [8].
In 2001 Givental [9] has announced a proof of Virasoro conjecture forPn. Y.-P. Lee and R. Pandharipande are writing a book [16] giving details.
Recently Teleman [23] announced a proof of the Virasoro conjecture for manifolds with semi-simple quantum cohomology. His argument depends crucially on the Mumford conjecture about the stable rational cohomology rings of the moduli spaces proved by Madsen and Weiss [20].
Goulden, Jackson and Vakil [10] recently give an enlightening proof of identity (1) for up to three points. Their remarkable proof uses relative virtual localization and a combinatorialization of the Hodge integrals, establishing connections to double Hurwitz numbers.
Our alternative approach is quite direct, we prove identity (3) for all g and n by using a recursive formula of n-point functions. Actually, the n-point function formula has far-reaching applications. Recently Zhou [28] used our results on n-point functions in his computation of Hurwitz-Hodge integrals, which leads to a proof of the crepant resolution conjecture of type A surface singularities for all genera.
Acknowledgements. The authors would like to thank Professors Sergei Lando, Jun Li, Chiu-Chu Melissa Liu, Ravi Vakil and Jian Zhou for helpful communications.
2. The n-point functions
Definition 2.1. We call the following generating function F(x1, . . . , xn) =
∞
X
g=0
Fg(x1, . . . , xn)
=
∞
X
g=0
X
Pdj=3g−3+n
hτd1· · ·τdnig n
Y
j=1
xdjj
then-point functions.
Consider the following “normalized” n-point function G(x1, . . . , xn) = exp −Pn
j=1x3j 24
!
F(x1, . . . , xn).
We will let Gg(x1, . . . , xn) denote the degree 3g−3 +n homogenous component ofG(x1, . . . , xn).
In contrast with the original n-point function, its normalization has some distinct properties (see [18]). For example, the coefficient of zkQn
j=1xdjj inGg(z, x1, . . . , xn) is zero wheneverk >2g−2 +n.
It’s well-known that
F0(x1, . . . , xn) =G0(x1, . . . , xn) = (x1+· · ·+xn)n−3.
There are explicit formulae for one and two-point functions due to Witten [25] and Dijkgraaf (see [2]) respectively
G(x) = 1
x2, G(x, y) = 1 x+y
X
k≥0
k!
(2k+ 1)!
1
2xy(x+y) k
. In an unpublished note [27] (kindly sent to us by Faber), Zagier obtained a marvelous formula of the three-point function (see [18]).
We proved in [18] the following recursion formula for general normal- ized n-point function.
Proposition 2.2. [18]For n≥2, G(x1, . . . , xn) = X
r,s≥0
(2r+n−3)!!
4s(2r+ 2s+n−1)!!Pr(x1, . . . , xn)∆(x1, . . . , xn)s, where Pr and∆ are homogeneous symmetric polynomials defined by
∆(x1, . . . , xn) = Pn
j=1xj3
−Pn
j=1x3j
3 ,
Pr(x1, . . . , xn) =
1 2Pn
j=1xj X
n=I‘ J
X
i∈I
xi
2 X
i∈J
xi
2
G(xI)G(xJ)
3r+n−3
= 1
2Pn j=1xj
X
n=I‘ J
X
i∈I
xi2 X
i∈J
xi2 r
X
r′=0
Gr′(xI)Gr−r′(xJ), where I, J 6= ∅, n = {1,2, . . . , n} and Gg(xI) denotes the degree 3g+
|I| − 3 homogeneous component of the normalized |I|-point function G(xk1, . . . , xk|I|), where kj ∈I.
The proof amounts to check thatG(x1, . . . , xn), as recursively defined in Proposition 2.2, satisfies the following Witten-Kontsevich differential equation (see [18]),
y ∂
∂y
y+
n
X
j=1
xj2
Gg(y, x1, . . . , xn)
= y
8
y+
n
X
j=1
xj4
Gg−1(y, x1, . . . , xn)−y3 8
y+
n
X
j=1
xj2
Gg−1(y, x1, . . . , xn)
+y 2
X
n=I‘ J
y+X
i∈I
xi
! X
i∈J
xi
!3
+ 2 y+X
i∈I
xi
!2
X
i∈J
xi
!2
×Gg′(y, xI)Gg−g′(xJ)
−1 2
y+
n
X
j=1
xj
n
X
j=1
xj
Gg(y, x1, . . . , xn).
The verification is tedious but straightforward. It will be included in a updated version of the paper [18].
Recall the well-known string equation hτ0
n
Y
i=1
τkiig =
n
X
j=1
hτkj−1Y
i6=j
τkiig
and the dilaton equation hτ1
n
Y
i=1
τkiig = (2g−2 +n)h
n
Y
i=1
τkiig.
Note that the string equation can be equivalently written as F(x1, . . . , xn,0) =Xn
j=1
xj
F(x1, . . . , xn).
Proposition 2.3. Let n ≥ 2. We have the following recursive for- mula of n-point functions.
(2g+n−1)Fg(x1, . . . , xn) = Pn
j=1xj3
12 Fg−1(x1, . . . , xn)
+ 1
2 Pn
j=1xj
g
X
g′=0
X
n=I‘ J
X
i∈I
xi
!2
X
i∈J
xi
!2
Fg′(xI)Fg−g′(xJ).
Proof. From Proposition 2.2, we have Gg(x1, . . . , xn)
= X
r+s=g
(2r+n−3)!!
4s(2g+n−1)!!Pr(x1, . . . , xn)∆(x1, . . . , xn)s
= 1
2g+n−1Pg(x1, . . . , xn)
+ X
r+s=g−1
(2r+n−3)!!
4s+1(2g+n−1)!!Pr(x1, . . . , xn)∆(x1, . . . , xn)s+1
= 1
(2g+n−1)Pg(x1, . . . , xn) +∆(x1, . . . , xn)
4(2g+n−1)Gg−1(x1, . . . , xn).
We define H= exp
Pn i=1x3i
24
, H−1 = exp
−Pn i=1x3i 24
,
Hd= 1 d!
Pn i=1x3i
24 d
, Hd−1 = 1 d!
−Pn
i=1x3i 24
d
.
Note thatPd
i=0HiHd−i−1 = 0 if d >0.
Let LHS and RHS denote the left and right hand side of the recursion in the lemma. We have
H−1·RHS=
∞
X
g=0
1 12
n
X
i=1
xi
!3
Gg−1(x1, . . . , xn) +Pg(x1, . . . , xn)
=
∞
X
g=0
(2g+n−1)Gg(x1, . . . , xn) + 1 12
n
X
i=1
x3i
!
Gg−1(x1, . . . , xn)
!
H−1·LHS=
∞
X
g=0
X
a+b+c=g
(2a+ 2b+n−1)Ga(x1, . . . , xn)HbHc−1
=
∞
X
g=0 g
X
a=0
(2a+n−1)Ga(x1, . . . , xn) X
b+c=g−a
HbHc−1
+
∞
X
g=0
X
a+b+c=g
Ga(x1, . . . , xn)2bHbHc−1
=
∞
X
g=0
(2g+n−1)Gg(x1, . . . , xn) +
∞
X
g=0
1 12
n
X
i=1
x3i
!
Gg−1(x1, . . . , xn).
q.e.d.
Only very recently, we realize that Proposition 2.3 has already been embodied in the first KdV equation of the Witten-Kontsevich theorem.
The KdV hierarchy is the following hierarchy of differential equations forn≥1,
∂U
∂tn = ∂
∂t0Rn+1,
where Rn are Gelfand-Dikii differential polynomials in U, ∂U/∂t0,
∂2U/∂t20, . . ., defined recursively by R1=U, ∂Rn+1
∂t0 = 1
2n+ 1 ∂U
∂t0Rn+ 2U∂Rn
∂t0 + 1 4
∂3
∂t30Rn
. It is easy to see that
R2 = 1
2U2+ 1 12
∂2U
∂t20 , R3 = 1
6U3+ U 12
∂3U
∂t30 + 1 24
∂U
∂t0 2
+ 1 240
∂4U
∂t40 , ...
The Witten-Kontsevich theorem states that the generating function F(t0, t1, . . .) =X
g
X
n
h
∞
Y
i=0
τiniig
∞
Y
i=0
tnii ni!
is a τ-function for the KdV hierarchy, i.e. ∂2F/∂t20 obeys all equations in the KdV hierarchy. The first equation in the KdV hierarchy is the classical KdV equation
∂U
∂t1 =U∂U
∂t0 + 1 12
∂3U
∂t30 .
By the Witten-Kontsevich theorem, we have
∂3F
∂t1∂t20 = ∂F
∂t20
∂F
∂t30 + 1 12
∂5F
∂t50 .
Integrating each side with respect to t0 and putting hhτk1· · ·τknii:=
∂nF/∂tk1· · ·∂tkn, we get hhτ0τ1ii= 1
12hhτ04ii+1
2hhτ02iihhτ02ii.
Then Proposition 2.3 follows by applying the dilaton equation.
3. The Faber intersection number conjecture
Now we explain our approach to prove identity (3), hence the Faber intersection number conjecture. We establish its relationship with n- point functions.
For the sake of brevity, we introduce the following notations La,bg (y, x1. . . , xn)
=
g
X
g′=0
X
n=I‘ J
y+X
i∈I
xi
a
−y+X
i∈J
xi
b
Fg′(y, xI)Fg−g′(−y, xJ), where a, b ∈ Z. We regard La,bg (y, x1. . . , xn) as a formal series in Q[x1, . . . , xn][[y, y−1]] with degy <∞.
We now prove that the Faber intersection number conjecture can be reduced to three statements about the coefficients of the above func- tions.
Proposition 3.1. We have i)
L0,0g (y, x1. . . , xn)
y2g−2 = 0;
ii) For k >2g,
L2,2g (y, x1. . . , xn)
yk = 0;
iii) For dj ≥1 andPn
j=1dj =g+n, L2,2g (y, x1. . . , xn)
y2gQn
j=1xdjj = (2g+n+ 1)!
4g(2g+ 1)!Qn
j=1(2dj−1)!!. In fact, Proposition 3.1 is a special case of more general results proved in the next section. Clearly identities (i) and (ii) of the following corol- lary add up to the desired identity (3).
Corollary 3.2. We have
i) Let dj ≥0 and Pn
j=1dj =g+n−2. Then h
n
Y
j=1
τdjτ2gig =
n
X
j=1
hτdj+2g−1
Y
i6=j
τdiig
− 1 2
X
n=I‘ J
2g−2
X
j=0
(−1)jhτj
Y
i∈I
τdiig′hτ2g−2−j
Y
i∈J
τdiig−g′; ii) Let dj ≥1 and Pn
j=1(dj−1) =g−1. Then
2g
X
j=0
(−1)jhτ2g−jτj
n
Y
i=1
τdiig= (2g+n−1)!
4g(2g+ 1)!Qn
j=1(2dj−1)!!; iii) Let k > g, dj ≥0 and Pn
j=1dj = 3g+n−2k−2. Then
2k
X
j=0
(−1)jhτ2k−jτj
n
Y
i=1
τdiig = 0.
Proof. Since one and two-point functions in genus 0 are F0(x) = 1
x2, F0(x, y) = 1 x+y =
∞
X
k=0
(−1)k xk yk+1, it is consistent to define
hτ−2i0 = 1, hτkτ−1−ki0 = (−1)k, k ≥0.
By allowing the index to run over all integers, we have
1 2
X
n=I‘ J
2g−2
X
j=0
(−1)jhτjY
i∈I
τdiig′hτ2g−2−jY
i∈J
τdiig−g′
+h
n
Y
j=1
τdjτ2gig−
n
X
j=1
hτdj+2g−1Y
i6=j
τdiig
= 1 2
X
n=I‘ J
X
j∈Z
(−1)jhτj
Y
i∈I
τdiig′hτ2g−2−j
Y
i∈J
τdiig−g′
=
g
X
g′=0
X
n=I‘ J
Fg′(y, xI)Fg−g′(−y, xJ)
y2g−2Qn i=1xdii
=
L0,0g (y, x1, . . . , xn)
y2g−2Qn
i=1xdii = 0.
From Proposition 2.3, we have 1
2
n
X
j=1
xj
Fg(x1, . . . , xn) = Pn
j=1xj4
24(2g+n−1)Fg−1(x1, . . . , xn)
+ 1
2(2g+n−1)
L2,2g (y, xn) +
g
X
g′=0
X
n=I‘ J
X
i∈I
xi
!2
X
i∈J
xi
!2
×Fg′(y,−y, xI)Fg−g′(xJ)
! . By Proposition 3.1(ii)-(iii), we can use Proposition 2.3 to inductively prove
2k
X
j=0
(−1)jhτ2k−jτj
n
Y
i=1
τdiig= [Fg(y,−y, x1, . . . , xn)]y2k = 0, fork > g and we have
2g
X
j=0
(−1)jhτ2g−jτjτ0
n
Y
i=1
τdiig = (2g+n)!
4g(2g+ 1)!Qn
j=1(2dj−1)!!, which, from the string equation and induction on the maximum index (say d1) among {di}, implies (by the dilaton equation, we may assume di≥2)
2g
X
j=0
(−1)jhτ2g−jτj n
Y
i=1
τdiig
=
2g
X
j=0
(−1)jhτ0τ2g−jτjτd1+1
n
Y
i=2
τdiig
−
n
X
k=2 2g
X
j=0
(−1)jhτ2g−jτjτd1+1τdk−1 Y
i6=1,k
τdiig
= (2g+n)!
4g(2g+ 1)!Qn
j=1(2dj−1)!!(2d1 + 1)
−
n
X
k=2
(2g+n−1)!(2dk−1) 4g(2g+ 1)!Qn
j=1(2dj −1)!!(2d1 + 1)
= (2g+n−1)!
4g(2g+ 1)!Qn
j=1(2dj−1)!!.
q.e.d.
So in order to prove the Faber intersection number conjecture, we only need to prove the three statements (i)-(iii) in Proposition 3.1 about n- point functions. Actually we will prove more general results which are stated as main theorems, Theorems 4.4 and 4.5 in the next section.
Proposition 3.1, therefore the Faber intersection number conjecture, is a special case of these theorems.
4. Proof of main theorems The binomial coefficients kp
, fork≥0, p∈Z are given by p
k
=
0, k <0,
1, k= 0,
p(p−1)···(p−k+1)
k! , k≥1.
Lemma 4.1. Let a, b∈Z andn≥0. Then
n
X
i=0
i+a i
n−i+b n−i
=
n+a+b+ 1 n
. Proof. Note that
p k
=
p−1 k
+
p−1 k−1
.
By denoting the left-hand side of the above equation by An(a, b), we have
An(a, b) =An(a−1, b) +An−1(a, b).
First we argue by induction on nand |b| to prove An(0, b) =
n+b+ 1 n
.
Then we argue by induction on nand |a|to prove An(a, b) =
n+a+b+ 1 n
.
q.e.d.
We now prove two lemmas that will serve as base cases for our in- ductive arguments.
Lemma 4.2. Let a, b∈Z andk≥2g−3 +a+b. Then i)
h
La,bg (y, x)i
yk = 0,
ii)
h
La,bg (y, x)i
y2g−4+a+bxg+1 = (−1)b(2g−2 +a+b) 4g(2g+ 1)!! .
Proof. Here we recall the definition of normalizedn-point functions G(x1, . . . , xn) = exp −Pn
j=1x3j 24
!
·F(x1, . . . , xn).
In particular, we have G(x) = 1
x2, G(x, y) = 1 x+y
X
k≥0
k!
(2k+ 1)!
1
2xy(x+y) k
. By definition
X
g≥0
La,bg (y, x1. . . , xn) =
exp Pn
j=1x3j 24
! X
n=I‘ J
y+X
i∈I
xia
−y+X
i∈J
xib
G(y, xI)G(−y, xJ), So for statements (i) and (ii), it is not difficult to see that we only need to prove that for k≥2g−3 +a+b,
h
ya−2(−y+x)bGg(−y, x) + (−y)b−2(y+x)aGg(y, x)i
yk = 0 and
h
ya−2(−y+x)bGg(−y, x) + (−y)b−2(y+x)aGg(y, x)i
y2g−4+a+bxg+1
= (−1)b(2g−2 +a+b) 4g(2g+ 1)!! .
Both follow easily from the explicit formula ofG(y, x). q.e.d.
Lemma 4.3. Let a, b∈Z andk≥a+b−3. Then i)
hLa,b0 (y, x1, . . . , xn)i
yk = 0, ii)
h
La,b0 (y, x1, . . . , xn)i
ya+b−4Qn j=1xj
= (−1)b(a+b+n−3)!
(a+b−3)! . Proof. Since
F0(x1, . . . , xn) = (x1+· · ·+xn)n−3, we have by definition
La,b0 (y, x1. . . , xn) = X
n=I‘ J
y+X
i∈I
xi|I|−2+a
−y+X
i∈J
xi|J|−2+b
.
For any monomial ykQn
j=1xdjj in La,b0 (y, x1. . . , xn), if k ≥ a+b−3, then there must be somedj = 0. We may assumedn= 0, then
La,b0 (y, x1. . . , xn−1,0)
= X
{1,...,n−1}=I‘ J
y+X
i∈I
xi|I|−1+a
−y+X
i∈J
xi|J|−2+b
+
y+X
i∈I
xi|I|−2+a
−y+X
i∈J
xi|J|−1+b!
=
n−1
X
j=1
xj
X
{1,...,n−1}=I‘ J
x1+X
i∈I
xi|I|−2+a
−x1+X
i∈J
xi|J|−2+b
=
n−1
X
j=1
xj
La,b0 (y, x1. . . , xn−1).
So (i) follows by induction onn. By applying Lemma 4.1 we have
hLa,b0 (y, x1, . . . , xn)i
ya+b−4Qn j=1xj
= (−1)b
n
X
|I|=0
|I| −2 +a
|I|
|I|!
|j| −2 +b
|J|
|J|!
n
|I|
= (−1)bn!
n
X
i=0
i−2 +a i
n−i−2 +b n−i
= (−1)bn!
a+b+n−3 n
= (−1)b(a+b+n−3)!
(a+b−3)! .
So we have proved (ii). q.e.d.
Theorem 4.4. Let a, b∈Z and k≥2g−3 +a+b. Then h
La,bg (y, x1. . . , xn)i
yk = 0.
Proof. We will argue by induction ongandn, since the theorem holds forg= 0 or n= 1 as proved in the above lemmas. We have
(2g+n)La,bg (y, x1. . . , xn)
=
g
X
g′=0
X
n=I‘ J
y+X
i∈I
xia
−y+X
i∈J
xib
(2g′+|I|)Fg′(y, xI)Fg−g′(−y, xJ) +
g
X
g′=0
X
n=I‘ J
y+X
i∈I
xia
−y+X
i∈J
xib
Fg′(y, xI)(2g−2g′+|J|)Fg−g′(−y, xJ).
SubstitutingFg′(y, xI) by Propostion 2.3,
g
X
g′=0
X
n=I‘ J
y+X
i∈I
xia
−y+X
i∈J
xib
(2g′ +|I|)
×Fg′(y, xI)Fg−g′(−y, xJ)
#
yk
= 1 12
hLa+3,bg−1 (y, x1, . . . , xn)i
yk
+
g
X
g′=0
X
s≥0
a−1 s
X
n=I‘ J
Fg′(xI) X
i∈I
xis+2
La+1−s,bg−g′ (y, xJ)
yk
.
Note that in the last term of the above equation, |J| < n. So by induction, fork≥2g−3 +a+b, the sums vanish except for g′ = 0 and s= 0, namely the term
X
n=I‘ J
X
i∈I
xi
|I|−1
La+1,bg (y, xJ)
yk
.
Letdj ≥1 for 1≤j≤n. By induction, it is not difficult to see from the above that
(2g+n)h
La,bg (y, x1. . . , xn)i
ykQn j=1xdjj
= 1 12
h
La+3,bg−1 (y, x1, . . . , xn) +La,b+3g−1 (y, x1, . . . , xn)i
ykQn j=1xdjj .
By induction, we have
0 =
n
X
j=1
xj
h
La+1,b+1g−1 (y, x1, . . . , xn)i
yk fork≥2g−3 +a+b
=h
La+2,b+1g−1 (y, x1, . . . , xn) +La+1,b+2g−1 (y, x1, . . . , xn)i
yk
and
0 =
n
X
j=1
xj
3
h
La,bg−1(y, x1, . . . , xn)i
yk fork≥2g−5 +a+b
=h
La+3,bg−1 (y, x1, . . . , xn) +La,b+3g−1 (y, x1, . . . , xn)i
yk
+ 3h
La+2,b+1g−1 (y, x1, . . . , xn) +La+1,b+2g−1 (y, x1, . . . , xn)i
yk
=h
La+3,bg−1 (y, x1, . . . , xn) +La,b+3g−1 (y, x1, . . . , xn)i
yk.
So we have proved that h
La,bg (y, x1. . . , xn)i
ykQn
j=1xdjj = 0, fordj ≥1.
If some dj is zero, the above identity still holds by applying the string equation
La,bg (y, x1. . . , xn,0) =
n
X
j=1
xj
La,bg (y, x1. . . , xn).
So we proved the theorem. q.e.d.
Theorem 4.5. Let a, b∈Z, dj ≥1 and P
jdj =g+n. Then h
La,bg (y, x1. . . , xn)i
y2g−4+a+bQn j=1xdjj
= (−1)b(2g−3 +n+a+b)!
4g(2g−3 +a+b)!Qn
j=1(2dj−1)!!.
Proof. By the dilaton equation, we may assume dj ≥ 2. As in the proof of the above theorem, we have
(2g+n)h
La,bg (y, x1. . . , xn)i
y2g−4+a+bQn j=1xdjj
= 1 12
hLa+3,bg−1 (y, xn) +La,b+3g−1 (y, xn)i
y2g−4+a+bQn j=1xdjj
=−1 4
h
La+2,b+1g−1 (y, xn) +La+1,b+2g−1 (y, xn)i
y2g−4+a+bQn j=1xdjj
=−1 4
" n X
i=1
xi
!
La+1,b+1g−1 (y, xn)
#
y2g−4+a+bQn j=1xdjj
=−1 4
n
X
j=1
h
La+1,b+1g−1 (y, xn)i
y2g−4+a+bxdjj −1Q
i6=jxdii
= (−1)b(2g−3 +n+a+b)!
4g(2g−3 +a+b)!Qn
j=1(2dj−1)!!
n
X
j=1
(2dj−1)
= (2g+n) (−1)b(2g−3 +n+a+b)!
4g(2g−3 +a+b)!Qn
j=1(2dj−1)!!.
So we have proved the theorem. q.e.d.
All the three statements in Proposition 3.1 are particular cases of Theorems 2.4 and 2.5. We thus conclude the proof of the Faber inter- section number conjecture.
The following corollaries ware stated as conjectures in our previous paper [17].
Corollary 4.6. Let dj ≥1 and Pn
j=1(dj−1) =g. Then (2g−3 +n)!
22g+1(2g−3)!Qn
j=1(2dj−1)!!
=hτ2g−2 n
Y
j=1
τdjig−
n
X
j=1
hτdj+2g−3
Y
i6=j
τdiig
+ 1 2
X
n=I‘ J
2g−4
X
j=0
(−1)jhτj
Y
i∈I
τdiig′hτ2g−4−j
Y
i∈J
τdiig−g′.
Proof. Since the right hand side is just 1
2
L0,0g (y, x1. . . , xn)
y2g−4Qn j=1xdjj ,
the result follows from Theorem 4.5. q.e.d.
Corollary 4.7. Let g≥2, dj ≥1 and Pn
j=1(dj−1) =g. Then
−(2g−2)!
|B2g−2| Z
Mg,n
ψd11· · ·ψdnnch2g−3(E)
= 2g−2
|B2g−2| Z
Mg,n
ψd11· · ·ψdnnλg−1λg−2−3 Z
Mg,n
ψ1d1· · ·ψdnnλg−3λg
!
= 1 2
2g−4
X
j=0
(−1)jhτ2g−4−jτjτd1· · ·τdnig−1
+ (2g−3 +n)!
22g+1(2g−3)!· 1 Qn
j=1(2dj −1)!!. Proof. We apply Mumford’s formulae [22]
(2g−3)!·ch2g−3(E) = (−1)g−1(3λg−3λg−λg−1λg−2),
ch2g−3(E) = B2g−2 (2g−2)!
"
κ2g−3−
n
X
i=1
ψ2g−3i
+1 2
X
ξ∈∆
lξ∗
2g−4
X
i=0
ψin+1(−ψn+2)2g−4−i
! # .
So the identity follows from Corollary 4.6. q.e.d.
Both Theorems 4.4 and 4.5 can be extended without difficulty.
Let us use the notation
Lg(y, za, wb, xn)
=
g
X
g′=0
X
n=I‘ J
Fg′(y, z1, . . . , za, xI)Fg−g′(−y, w1, . . . , wb, xJ).
Theorem 4.8. Let a≥0, b≥0, n≥1. We have i) For k≥2g−3 +a+b,
Lg(y, za, wb, xn)
yk = 0.
ii) For rj ≥0, sj ≥0, dj ≥1 and P
rj+P
sj+P
dj =g+n, Lg(y, za, wb, xn)
y2g−4+a+bQa
j=1zjrjQb
j=1wsjj Qn j=1xdjj
= 1
Qa
j=1(2rj+ 1)!!Qb
j=1(2sj+ 1)!!
· (−1)b(2g−3 +n+a+b)!
4g(2g−3 +a+b)!Qn
j=1(2dj −1)!!.
iii) For rj ≥0, sj ≥0, dj ≥1, P
rj+P
sj +P
dj =g+n+ 1 and u,#{rj = 0}, v,#{sj = 0}, w,#{dj = 1},
Lg(y, za, wb, xn)
y2g−5+a+bQa
j=1zjrjQb
j=1wsjj Qn
j=1xdjj = C
Qa
j=1(2rj+ 1)!!Qb
j=1(2sj+ 1)!!· (−1)b(2g−3 +n+a+b)!
4g(2g−4 +a+b)!Qn
j=1(2dj −1)!!, where the constant C is given by
C,
a
X
j=1
rj−
b
X
j=1
sj+a−b
2 + (5−u)u−(5−v)v 2(2g+n+a+b−3−w). Proof. Wheng= 0, the proof is an easy verification. Let p, q∈Z.
Lp,qg (y, za, wb, xn)
=
g
X
g′=0
X
n=I‘ J
(y+
a
X
i=1
zi+X
i∈I
xi)p(−y+
b
X
i=1
wi+X
i∈J
xi)q
×Fg′(y, z1, . . . , za, xI)Fg−g′(−y, w1, . . . , wb, xJ).
Exactly the same argument of Theorem 4.4 will prove that for k ≥ 2g−3 +p+q+a+b,
[Lp,qg (y, za, wb, xn)]yk = 0.
Statements (ii) and (iii) can also be proved similarly as Theorem 4.5.
q.e.d.
Theorem 4.8 proves all conjectures in Section 3 of [17]. We may write down the coefficients of Lg(y, za, wb, xn) explicitly to get a lot of interesting identities of intersection numbers. For example, when a= 1, b= 0,
Lg(y, z, xn)
ykzrQn j=1xdjj
= X
n=I‘ J
k
X
j=0
(−1)jhτj
Y
i∈I
τdiig′hτk−jτrY
i∈J
τdiig−g′
+hτk+2τr
n
Y
j=1
τdjig−(−1)khτk+r+1 n
Y
j=1
τdjig−
n
X
j=1
hτrτdj+k+1Y
i6=j
τdiig.